What are three methods of solving simultaneous equations?

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Simultaneous equations are two or more mathematical statements that are true at the same time and contain two unknowns. For example, it is possible to make two simultaneous equations regarding the cost of apples and bananas if one person bought three apples and two bananas for $2.80 while another person bought one apple and four bananas for $2.60. The equations would be 3a (apples) + 2b (bananas) = 280 and 1a + 4b = 260. In both equations, “a” and “b” will have the same value. Taken individually, each equation could have many different solutions for the value of “a” and “b.” However, there is only one correct answer when both equations are considered together, but it is impossible to solve definitively for two unknowns without using a method to simplify the calculation.

One method of solving simultaneous equations is called elimination. This can be done without changing the equations if the coefficient (the number preceding the letter) on one side of both equations is equal. If not, this method involves manipulating the equations through multiplication so that the coefficients on one side of both equations do equal each other. Then, one equation can be added or subtracted from the other, leaving one unknown rather than two.

To use the above example, multiplying the first equation by 2 on both sides will yield the equation 6a + 4b = 560. Because the second equation also has 4b, when it is subtracted from the altered first equation, the “c” variables can be eliminated. What remains is simple to solve: 5a = 300, meaning a = 60. It is now clear that apples cost 60 cents each, and the price of bananas can be easily deduced by substituting 60 for “a” in either original equation: 1(60) + 4b = 260; 4b = 200; b = 50. In other words, a banana costs 50 cents.

The elimination method works like this when there is already a matching coefficient: If one person bought three apples and two bananas for $2.80 and another person bought four apples and two bananas for $3.40, the two equations can be expressed as 3a + 2b = 280 and 4a + 2b = 340. Each equation already has a “2b.” Now the first equation can simply be subtracted from the second, resulting in 1a = 60, meaning one apple costs 60 cents. Now, to figure out the value of “b,” or how much each banana costs, simply substitute the value of “a” in either equation: 3(60) + 2b = 280; 180 + 2b = 280; 2b = 100; b = 50.

A second method for solving simultaneous equations is called substitution. To do this, one variable must be changed into x = something or y = something, and then substitute x or y into the other equation. This method isolates one of the variables. To use the original apple and banana example, the second equation can be rearranged to read a = 260 – 4b. This new definition of “a” can now be substituted into the first equation as follows: 3(260 – 4b) + 2b = 280; 780 – 12b + 2b = 280; 10b = 500; b = 50. Substitution is the better method to use when dealing with difficult simultaneous equations.

A third method is called graphing. In this method, each equation is mapped out on a graph where each line is a continuum of points representing possible pairs of solutions for each unknown. There will be only one point at which both lines intersect. This method is best for figuring out the unknowns in two simultaneous equations, rather than multiple simultaneous equations.

Resource by Will

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